Let $S_\infty$ considered as Polish Group. Prove that every Polish subgroup of $S_\infty$ has the following form: $\overline{{\left \langle X \right \rangle}}$, where $X$ is a countable subset of $S_\infty$.
I tried using the known facts that every Polish subspace is a $G_\delta$ set and that the closure of an intersection is contained in the intersection of the closures, but without success.
Thank you
The key point is to show that a Polish subgroup of a Polish group is closed.
Thus let $H \subset S_{\infty}$ be a Polish subgroup, and let $K = \overline{H}$. The closure of a subgroup is a subgroup in every topological group, so $K$ is a closed subgroup of $S_{\infty}$. In metrisable spaces, every closed set is a $G_{\delta}$, so $K$ is Polish. Since $H$ is a Polish subspace of $K$, it is a $G_{\delta}$ in $K$. Polish spaces are Baire spaces, and $H$ is dense in $K$, therefore $K \setminus H$ is meagre. But $K\setminus H$ is the union of cosets of $H$ is $K$, and translations are homeomorphisms, so if $K \setminus H$ were nonempty, $H$ itself would be meagre (as a subset of a meagre subset of $K$). But then $K = H \cup (K\setminus H)$ would be meagre, contradicting Baire's theorem. It follows that $K = H$, i.e. $H$ is closed.
And as a Polish space, $H$ is separable. So let $X$ be a countable dense subset of $H$, then $H = \overline{\langle X\rangle}$.